Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 39 1 Solution Created 2026-10-03 Updated 2026-10-07
Write and for the expectation and empirical measure. A sufficient bracketing of a function class condition is that, for every , finitely many brackets cover , with measurable integrable endpoints and . The uniform strong law from finite L1 bracketing then givesHere the observations are independent and identically distributed. For an uncountable class, one either assumes a measurable supremum, for example through a pointwise separable function class, or formulates the conclusion as a pathwise bound on a common probability-one event.
For the parameterized class, the Heine-Borel theorem makes compact. Put . A countable dense subset of gives the same supremum because of continuity, so is measurable. Define the modulusThe supremum is measurable by taking a countable dense subset of the compact set of admissible pairs. For each , uniform continuity on gives . Also , and . Thus the dominated convergence theorem gives .
Choose a finite net of radius in . The lower and upper envelopes of over each closed ball are measurable integrable function brackets; the same countable-dense-set argument applies within each such compact ball. Their widths are at most . They cover the whole class, so its bracketing of a function class condition follows. Moreover, dominated convergence shows that is continuous. Both empirical and population functions therefore have a supremum over a common countable dense parameter set. Applying the uniform strong law from finite L1 bracketing proves uniform almost-sure convergence over the entire compact parameter set, rather than merely convergence at each fixed parameter.
For the exponential family,where and . Thus a weak sufficient condition is bounded on , almost surely, and integrability of under the sampling law. No positive lower bound on over the whole real line is required. Zeros off the sampling support are harmless: choose arbitrary finite versions of the log-densities on that common null set when applying the function-class theorem.
If the sampling law is , one convenient assumption is that is a compact subset of the interior of the finite domain of the cumulant function of an exponential family, andOn that interior, is continuous, hence bounded on . Finiteness of at for some implies , hence . The displayed integral gives the remaining integrability. If the observations come from an arbitrary unrelated law, assumptions on and alone cannot control that law's tails; the sampling-law integrability must then be stated explicitly.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 33 1 Solution Created 2026-10-03 Updated 2026-10-07
Write and . A function bracket contains the measurable functions satisfying for every . Require its endpoints to be integrable and call its width. A sufficient condition is that, for every , finitely many brackets of width at most cover the whole class . Under this condition the uniform strong law from finite L1 bracketing statesThe conclusion holds outside a common measurable null set. If the supremum is not initially known to be measurable, this formulation means pathwise convergence on a measurable probability-one event; a pointwise separable function class, including the application below, has a measurable supremum. Pointwise brackets also ensure the sample inequalities hold simultaneously over the class.
To prove the result, choose a finite -cover , . If belongs to bracket , monotonicity of the empirical measure and of expectation givesandConsequentlyApply the strong law of large numbers to these finitely many integrable endpoints. On a probability-one event, the maximum tends to zero. Repeat with , , and intersect the countably many probability-one events. The limiting supremum is bounded by for every , hence is zero. This proves the uniform law of large numbers without a boundedness assumption on the class itself.
For the moment-generating function, use the empirical measure estimatorIf , this estimator and are both identically one. Otherwise, makes increasing in , and provides an integrable envelope. The dominated convergence theorem shows that is continuous on , hence uniformly continuous.
For any , choose a partition so that for every . If , then for every . These endpoint functions form finitely many integrable function brackets with the required widths. The just-proved uniform law of large numbers therefore givesBoth functions of are continuous; their supremum equals the supremum over a countable dense subset, so it is measurable. This proves uniform consistency of an empirical moment-generating function using only the observed sample.
Uniform strong law from finite L1 bracketing 2026-10-07
Suppose an integrable measurable-function class can be covered by finitely many function brackets of every positive width. For an independent sample with common law , the empirical measure satisfies on a common probability-one event. For one finite -cover, the supremum is bounded by plus the largest empirical error among its endpoints. The strong law of large numbers makes that finite maximum vanish. Taking a countable sequence of widths decreasing to zero proves the assertion. A measurable supremum can be obtained from a pointwise separable function class; otherwise the probability-one-event formulation expresses the same pathwise conclusion.